Question

use the references to access important values if needed for this question. a. how many atoms of nitrogen are present in 7.01 grams of dinitrogen tetrafluoride? atoms of nitrogen. b. how many grams of fluorine are present in 5.31×10²² molecules of dinitrogen tetrafluoride? grams of fluorine.

Explanation:

Step1: Determine the molar mass of $N_2F_4$

The molar mass of $N$ is approximately $14.01\ g/mol$ and of $F$ is approximately $19.00\ g/mol$. For $N_2F_4$, $M = 2\times14.01+4\times19.00=104.02\ g/mol$.

Step2: Calculate number of moles of $N_2F_4$ in part a

Given mass of $N_2F_4$ is $m = 7.01\ g$. Number of moles $n=\frac{m}{M}=\frac{7.01}{104.02}\ mol$.

Step3: Calculate number of nitrogen atoms in part a

In one molecule of $N_2F_4$, there are 2 nitrogen atoms. Using Avogadro's number $N_A = 6.022\times 10^{23}\ mol^{-1}$, the number of nitrogen atoms $N = 2\times n\times N_A=2\times\frac{7.01}{104.02}\times6.022\times 10^{23}\approx8.12\times 10^{22}$ atoms.

Step4: Calculate number of moles of $N_2F_4$ in part b

Number of molecules of $N_2F_4$ is $N_{molecules}=5.31\times 10^{22}$. Number of moles $n_{moles}=\frac{N_{molecules}}{N_A}=\frac{5.31\times 10^{22}}{6.022\times 10^{23}}\ mol$.

Step5: Calculate mass of fluorine in part b

In one molecule of $N_2F_4$, there are 4 fluorine atoms. Mass of fluorine $m_F=n_{moles}\times4\times19.00\ g/mol=\frac{5.31\times 10^{22}}{6.022\times 10^{23}}\times4\times19.00\approx6.71\ g$.

Answer:

a. $8.12\times 10^{22}$ atoms
b. $6.71$ g